How do you get this Equation in standard form?x²-3x-4=0?

x^2 - 3x - 4 = 0
x^2 + x - 4x - 4 = 0
(x^2 + x) - (4x + 4) = 0
x(x + 1) - 4(x + 1) = 0
(x + 1)(x - 4) = 0

x + 1 = 0
x = -1

x - 4 = 0
x = 4

∴ x = -1 , 4

x²-3x-4=0
This equation is already in standard form: ax² + bx + c = 0.
I think what you are being asked is to solve this equation.
You can do this either by factoring or by the quadratic formula.

Using factoring:
x² - 3x - 4 = 0; here you look for two integers whose sum is -3 and whose product is -4, such that when you do FOIL, the original equation will be the result; thus
(x + 1)(x - 4) = 0

Equate each factor to 0, then solve for x:
x + 1 = 0; therefore, one solution is x = -1.
x - 4 = 0; therefore, another solution is x = 4

The term 'standard form' refers to an equation with an x and a y term in it.To solve the equation. It takes the form Ax + By = C where A and B are integer coefficients of the x and y values.

[added]
As como said above, you can factor to get the answer more easily than using the quadratic formula below.


One way to solve this equation is to use the quadratic formula:

x = [-b +- sqrt(b^2 - 4ac)] / 2a using a = 1, b = -3, and c = -4


Plugging in those values:

x = [3 +- sqrt(-3^2 - 4 * 1 * (-4))] / 2 * 1

x = [3 +- sqrt(9 + 16)] / 2

x = [3 +- sqrt(25)] / 2

x = (3 + 5) / 2 = 8 / 2 = 4

and

x = (3 - 5) / 2 = -2 / 2 = -1

(x - 4) (x + 1) = 0
x = 4 , x = - 1